Q. No.Which one of the following compounds shows the presence of an intramolecular hydrogen bond?
1. H2O2
2. HCN
3. Cellulose
4. Concentrated acetic acid
The difference between amylose and amylopectin is:
| 1. | Amylopectin has 1→4 α - linkage and 1→6 α-linkage. |
| 2. | Amylose has 1→4 α-linkage and 1 → 6 β-linkage. |
| 3. | Amylopectin has 1 → 4 α-linkage and 1 → 6 β-linkage. |
| 4. | Amylose is made up of glucose and galactose. |
The central dogma of molecular genetics states that the genetic information flows from:
1. amino acid - proteins - DNA
2. DNA - carbohydrates - proteins
3. DNA - RNA - proteins
4. DNA - RNA - Carbohydrates
Which of the following statements is not correct?
| 1. | Ovalbumin is a simple food reserve in egg white. |
| 2. | Blood proteins thrombin and fibrinogen are involved in blood clotting. |
| 3. | Denaturation makes the proteins more active. |
| 4. | Insulin maintains sugar level in the blood of the human body. |
| 1. | β -glycosidic bond | 2. | Peptide bond |
| 3. | Dative bond | 4. | α -glycosidic bond |
| 1. | The sugar component in RNA is ribose and the sugar component in DNA is 2'-deoxyribose. |
| 2. | The sugar component in RNA is arabinose and the sugar component in DNA is ribose. |
| 3. | The sugar component in RNA is 2'-deoxyribose and the sugar component in DNA is arabinose. |
| 4. |
The sugar component in RNA is arabinose and the sugar component in DNA is 2’-deoxyribose.
|
D(+) glucose yields an oxime with hydroxyl amine. The structure of the oxime would be:
| 1. |  |
2. |  |
| 3. |  |
4. |  |
Sucrose can be formed by:
1. α–D–galactopyranose and α–D–glucopyranose
2. α–D–glucopyranose and β–D–fructofuranose
3. β–D–galactopyranose and α–D–fructofuranose
4. α–D–galactopyranose and β–D–fructopyranose
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